find the value of ∫_0 ^(π/2) ((xdx)/(1+cosx))


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Question Number 37071 by math khazana by abdo last updated on 08/Jun/18

$f i n d t h e v a l u e o f \int_{0}^{\frac{π}{2}} \frac{x d x}{1 + c o s x}$
Commented by math khazana by abdo last updated on 10/Jun/18
$changement tan((x/2))=t give I = ∫_0 ^1 ((2arctan(t))/(1+((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =4 ∫_0 ^1 ((arctan(t))/(1+t^2 +1−t^2 ))dt =2 ∫_0 ^1 arctan(t)dt and by parts I =2 { [t arctan(t)]_0 ^1 −∫_0 ^1 (t/(1+t^2 ))dt} =2{ (π/4) −(1/2)[ln(1+t^2 )]_0 ^1 } =(π/2) −ln(2) ★ I =(π/2) −ln(2)★$
$c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $I = \int_{0}^{1} \frac{2 a r c t a n (t)}{1 + \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = 4 \int_{0}^{1} \frac{a r c t a n (t)}{1 + t^{2} + 1 - t^{2}} d t$ $= 2 \int_{0}^{1} a r c t a n (t) d t a n d b y p a r t s$ $I = 2 {{[t a r c t a n (t)]}_{0}^{1} - \int_{0}^{1} \frac{t}{1 + t^{2}} d t}$ $= 2 {\frac{π}{4} - \frac{1}{2} {[l n (1 + t^{2})]}_{0}^{1}}$ $= \frac{π}{2} - l n (2)$ $★ I = \frac{π}{2} - l n (2) ★$
	Answered by tanmay.chaudhury50@gmail.com last updated on 09/Jun/18
	$∫_0 ^(Π/2) (x/(2cos^2 (x/2)))dx (1/2)∫_0 ^(Π/2) ((xsec^2 (x/2))/)dx let I_1 =∫xsec^2 (x/2)dx =x∫sec^2 (x/2)dx−∫[(dx/dx)∫sec^2 (x/2)dx]dx =x((tan(x/2))/(1/2))−∫((tan(x/2))/(1/2))dx =x×2tan(x/2)−4ln∣sec(x/2)∣ so required ans is (1/2){2xtan(x/2)−4ln∣sec(x/2)∣}_0 ^(Π/2) =(1/2)[{2×(Π/2)×1−4ln((√2) )}−{2×0×0−4ln1}] =(Π/2)−2×(1/2)ln2 =(Π/2)−ln2$
	$\int_{0}^{\frac{Π}{2}} \frac{x}{2 {c o s}^{2} \frac{x}{2}} d x$ $\frac{1}{2} \int_{0}^{\frac{Π}{2}} \frac{{x s e c}^{2} \frac{x}{2}}{} d x$ $l e t I_{1} = \int {x s e c}^{2} \frac{x}{2} d x$ $= x \int {s e c}^{2} \frac{x}{2} d x - \int [\frac{d x}{d x} \int {s e c}^{2} \frac{x}{2} d x] d x$ $= x \frac{t a n \frac{x}{2}}{\frac{1}{2}} - \int \frac{t a n \frac{x}{2}}{\frac{1}{2}} d x$ $= x \times 2 t a n \frac{x}{2} - 4 l n ∣ s e c \frac{x}{2} ∣$ $s o r e q u i r e d a n s i s$ $\frac{1}{2} {2 x t a n \frac{x}{2} - 4 l n ∣ s e c \frac{x}{2} ∣}_{0}^{\frac{Π}{2}}$ $= \frac{1}{2} [{2 \times \frac{Π}{2} \times 1 - 4 l n (\sqrt{2})} - {2 \times 0 \times 0 - 4 l n 1}]$ $= \frac{Π}{2} - 2 \times \frac{1}{2} l n 2$ $= \frac{Π}{2} - l n 2$
	Commented by math khazana by abdo last updated on 10/Jun/18

	$c o r r e c t a n s w e r t h a n k s s i r T a n m a y .$
	Answered by MJS last updated on 10/Jun/18

	$\int \frac{x}{1 + \cos x} d x =$ $[\begin{matrix} \int u^{'} v = u v - \int {u v}^{'} \\ u^{'} = \frac{1}{1 + \cos x} \Rightarrow u = \frac{\sin x}{1 + \cos x} \\ [\begin{matrix} u = \int \frac{d x}{1 + \cos x} = \int \frac{\tan \frac{x}{2}}{\sin x} d x = \\ [t = \frac{x}{2} \to d x = 2 d t] \\ = 2 \int \frac{\tan t}{\sin 2 t} d t = \int \frac{\tan t}{\sin t \cos t} d t = \int \sec^{2} t d t = \\ = \tan t = \tan \frac{x}{2} = \frac{\sin x}{1 + \cos x} \end{matrix}] \\ v = x \Rightarrow v^{'} = 1 \end{matrix}]$ $= \frac{x \sin x}{1 + \cos x} - \int \frac{\sin x}{1 + \cos x} d x =$ $[t = 1 + \cos x \to d x = - \frac{d t}{\sin x}]$ $= \frac{x \sin x}{1 + \cos x} + \int \frac{d t}{t} = \frac{x \sin x}{1 + \cos x} + \ln t =$ $= \frac{x \sin x}{1 + \cos x} + \ln (1 + \cos x) + C$
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